Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $${t_2}$$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $$\frac{{{t_1}}}{{{t_2}}}$$ will be

Solution :
Initial energy of capacitor, $${E_1} = \frac{{q_1^2}}{{2C}}$$
Final energy of capacitor, $${E_2} = \frac{1}{2}{E_1} = \frac{{q_1^2}}{{4C}} = {\left( {\frac{{\frac{{{q_1}}}{{\sqrt 2 }}}}{{2C}}} \right)^2}$$
$$\therefore {t_1}$$  = time for the charge to reduce to $$\frac{1}{{\sqrt 2 }}$$ of its initial value and $${t_2}$$ = time for the charge to reduce to $$\frac{1}{4}$$ of its initial value
$$\eqalign{ & {\text{We}}\,{\text{have,}}\,{q_2} = {q_1}{e^{ - \frac{t}{{CR}}}} \cr & \Rightarrow \ln \left( {\frac{{{q_2}}}{{{q_1}}}} \right) = - \frac{t}{{CR}}\quad \therefore \ln \left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{{ - {t_1}}}{{CR}}\,......\left( 1 \right) \cr & {\text{and}}\,\ln \left( {\frac{1}{4}} \right) = \frac{{ - {t_2}}}{{CR}}\,......\left( 2 \right) \cr} $$
By (1) and (2),
$$\frac{{{t_1}}}{{{t_2}}} = \frac{{\ln \left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\ln \left( {\frac{1}{4}} \right)}} = \frac{1}{2}\frac{{\ln \left( {\frac{1}{2}} \right)}}{{2\ln \left( {\frac{1}{2}} \right)}} = \frac{1}{4}$$